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trở thành một hằng đảng thức. Giá trị trong ô vuông là:
1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)
2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)
3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)
4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)
`a)x^3-8x^2+16x`
`=x(x^2-8x+16)`
`=x(x-4)^2`
`b)x^2+4y^2+2x-4y-4xy-24`
`=(x-2y)^2+2(x-2y)-24`
`=(x-2y)^2-4(x-2y)+6(x-2y)-24`
`=(x-2y-4)(x-2y+6)`
`c)x^4+x^3-x^2-2x-2`
`=x^4-2x^2+x^3-2x+x^2-2`
`=x^2(x^2-2)+x(x^2-2)+x^2-2`
`=(x^2-2)(x^2+x+1)`
a.
$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.
$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.
$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$
d.
$x^3-3x^2+3x-1=(x-1)^3$
e.
$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$
$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$
f.
$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$
\(a,x^3+6x^2y+12xy^2+8y^3\\ =x^3+3.2x^2+3.2^2.x+\left(2y\right)^3\\ =\left(x+2y\right)^3\)
\(b,x^3-3x^2+3x-1\\ =x^3-3x^2.1+3x.1^2-1^3\\ =\left(x-1\right)^3\)
a) \(x^3+6x^2y+12xy^2+8y^3\)
\(=x^3+3\cdot x^2\cdot2y+2\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)
\(=\left(x+2y\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
\(A=x^3-8-128-x^3=-136\\ B=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)
\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(128+x^3\right)=x^3-8-128-x^3=-136\)
\(B=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)
\(l,=5x\left(y^2-2yz+5z\right)\\ m,=\left(x+1\right)^3-27y^3\\ =\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\\ n,=\left(x-3y\right)^2\\ o,=\left(x+2y\right)^3\\ p,=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\\ q,=\left(x+2y\right)^2-2\left(x-2y\right)+1\\ =\left(x+2y-1\right)^2\)